Solving $xf^{\prime\prime}= 4f^{\prime}- 25x^9 f$

Question

Can someone explain to me how to solve the following differential equation, $$xf^{\prime\prime}= 4f^{\prime}- 25x^9 f \qquad \text{with initial conditions} \ f(0)=0, \ \ f^{'}(1)=1 $$ There is a hint which asks me to make the substitution, $t=x^5$. I really am a total novice in differential equations and my only attempt has been to write the function as a power series and try to guess the coefficients. It doesn't seem very inspiring though.

Answer 1

Substitute in the equation $$f'_x=\frac {df}{dx}=\frac {df}{dt}\frac {dt}{dx}=5x^4\frac {df}{dt}$$ $$f''_x=\frac {d^2f}{dx^2}=\frac {df}{dt}\frac {d^2t}{dx^2}+\frac {d^2f}{dt^2}(\frac {dt}{dx})^2$$ $$f''_x=\frac {20}{x^2} tf'_t+\frac {25}{x^2}t^2f''_t$$ Then the original equation becomes : $$f''_t+f_t=0$$ $$\implies r^2+1=0 \implies r=i,-i$$ $$f(t)=c_1\cos(t)+c_2\sin(t)$$ Substitute back $t=x^5$ $$f(x)=c_1\cos(x^5)+c_2\sin(x^5)$$ Apply initial conditions to get $c_1,c_2$

$$f(0)=0 \implies c_1=0 \implies f(x)=c_2\sin(x^5)$$ $$f'(x)=5c_2x^4\cos(x^5) $$ $$f'(1)=1 \implies 1=5c_2\cos(1) \implies c_2=\frac 15 \sec(1)$$ Therefore $$\boxed{f(x)=\frac 15 \sec(1)\sin(x^5)}$$